# NEET Questions Solved

If NaCl is doped with 10-4 mol % of SrCl2, the concentration of cation vacancies will be (NA = 6.023 x 1023 mol-1)

(a) 6.023 x 1015 mol-1                                            (b) 6.023 x 1016 mol-1

(c) 6.023 x 1017 mol-1                                            (d) 6.023 x 1014 mol-1

(c) Doping of NaCl with 10-4 mol % of SrCl2 means, 100 moles of NaCl are doped with 10-4  mol of SrCl2.

... 1 mol of NaCl is doped with

SrCl2 = 10-4/100 = 10-6 mole

As each Sr2+ ion introduces one cation vacancy.

... Concentration of cation vacancies

= 10-6 mol/mol of NaCl

= 10-6 x 6.023 x 1023 mol-1

= 6.023 x 1017 mol-1

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Metals have conductivity of the order of (ohm-1 cm-1):-

(a)1012
(b)108
(c)102
(d)10-6

(b) The conductance order of metals is 106 to 108 ohm-1 cm-1.

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A solid PQ have rock salt type structure in which Q atoms are at the corners of the unit cell. If the body centred atoms in all the units cells are missing, the resulting stoichiometry will be

(A) PQ

(B) PQ2

(C) P3Q4

(D) P4Q3

(C) In rock salt structure, atoms at the corner are also present at the face centres and hence

ZQ=8 x 1/8 + 6 x 1/2 =4

And, the atoms at body centre also present at edge centres. But from question, body centres are missing and hence

ZP = 12 x 1/4 = 3

Formula of resulting solid - P3Q4.

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Lithium borohydride (LiBH4), crystallises in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are:- a = 6.81 Å , b = 4.43 Å ,c = 717 Å. If the molar mass of LiBH4 is
21.76 g mol-1.The density of the crystal is:-

(A) 0.668 gcm-3
(B) 0.585 gcm2
(C) 1.23 gcm-3

(D) None of these

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A solid has a bcc structure. If the distance of closest approach between the two atoms is 1.73Å . The edge length of the cell is:

(a) 200pm

(b) $\sqrt{3}}{\sqrt{2}}$ pm

(c) 142.2 pm

(d) $\sqrt{2}$pm

(a) ratom$\sqrt{3}$/4a; Also closest approach in bcc lattice is 1/2 of body diagonal, i.e, $\sqrt{3}$/4a=1.73Å

or a = 1.73x2/ $\sqrt{3}$=1.996 Å=199.6pm

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The anions (A) form hexagonal closest packing and atoms (C) occupy only 2/3 of octahedral voids in
it. then the general formula of the compound is-

(A) CA

(B) A2

(C) C2A3

(D) C3A2

C) Hexagonal close packing contains 6 atoms per unit cell and hence the number of octahedral voids per unit cell is 6.
Hence, number of A atoms per unit cell=6 and number of C atoms per unit cell=6 x 2/3=4
The formula of ionic compound is given as simplest formula and hence formula is C2A3

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How many unit cells are present in a cube shaped ideal crystal of NaCl of mass 1.00g?

[ Atomic masses Na=23, Cl = 35.5]

(a) 2.57 x 1021

(b) 5.14 x 1021

(c) 1.28 x 1021

(d) 1.71 x 1021

(a) Mass of one unit cell = V x d (V is volume, d is density)

= a3 x d

Also, density = (n x at. mass) /(a3 x av.no.)

Mass of one unit cell = (a3 x n x at.mass) / (a3 x av.no.) = (n x at. mass )/(av.no.)

= 4x58.5 / 6.02 x 1023 (n=4 for cubic shape)

=38.87 x 10-23g

No. of unit cell in 1g = 1/38.87x10-23 = 2.57 x 1021

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The most efficient packing of similar spheres is obtained in

1. the simple cubic system and the body centered cubic system

2. the simple cubic system and the hexagonal close packed system

3. the face centered cubic system and the hexagonal close packed system

4. the body centered cubic system and the face centered cubic system

(3) The hexagonal close packing and the face-centred cubic system have the closest packing as their packing efficiency is 74%.

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In a compound, atoms of element Y form ccp lattice and those of element X occupy 2/3rd of tetrahedral voids. The formula of the compound will be :-

(a) X3Y4
(b)X4Y3
(c) X2Y3
(d) X2Y

(b) In a ccp array number of tetrahedral voids is twice the number of atoms.

x:y::1:2/3 x 2=3:4

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Close packing is maximum in the crystal lattice of:

(a) simple cubic
(b) face centred
(c) body centred
(d) none of these

(b) The maximum packing or the maximum proportion of volume filled by hard spheres in various arrangements are:

(a) simple cubic=π/6=0.52

(b) bcc=π/√3/8=0.68

(c) fcc=π/√2/6=0.74

(d) hep=π/√2/6=0.74

(e) Diamond=π/√3/6=0.34

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